SOLUTION
Given the values in the table, the following are the solutions to the questions.
Question A: SSx, SSy and SSxy
To get SSx:
[tex]\begin{gathered} SS_x=\sum ^{}_{}(x-\bar{x})^2=\sum ^{}_{}x^2-\frac{(\sum^{}_{}x)^2}{n} \\ \sum ^{}_{}x^2=1+4+9+16+25=55_{} \\ (\sum ^{}_{}x)^2=(1+2+3+4+5)^2=225_{} \\ n=5 \\ SS_x=\sum ^{}_{}x^2-\frac{(\sum^{}_{}x)^2}{n}=55-\frac{225}{5} \\ =55-45=10 \end{gathered}[/tex]
To get SSy:
[tex]\begin{gathered} SS_y=\sum ^{}_{}(y-\bar{y})^2=\sum ^{}_{}y^2-\frac{(\sum ^{}_{}y)^2}{n} \\ (\sum ^{}_{}y)^2=(14+11+9+9+6)^2=49^2=2401 \\ \sum ^{}_{}y^2=(196+121+81+81+36)=515 \\ n=5 \\ SS_y=515-\frac{2401}{5}=515-480.2=34.8 \end{gathered}[/tex]
To get SSxy:
[tex]\begin{gathered} SS_{xy}=\sum ^{}_{}(x-\bar{x})(y-\bar{y})=\sum ^{}_{}xy-\frac{(\sum ^{}_{}x)(\sum ^{}_{}y)}{n} \\ \sum ^{}_{}xy=(14+22+27+36+30)=129 \\ \sum ^{}_{}x=15 \\ \sum ^{}_{}y=49 \\ SS_{xy}=129-\frac{(15)(49)}{5}=129-147=-18 \end{gathered}[/tex]
Question b: Calculate r
[tex]\begin{gathered} r=\frac{SS_{xy}}{\sqrt[]{SS_x.SS_y}} \\ r=-\frac{18}{\sqrt[]{(10)(34.8)}}=\frac{-18}{\sqrt[]{348}}=-0.9649 \end{gathered}[/tex]
Given the value gotten above, I will need to calculate the critical value using the critical value table to ascertain if there is enough evidence.
[tex]\alpha=0.05,degree\text{ of fre}edom=n-2=5-2=3[/tex]
Using the critical value table to get the value at significance level 0.05 and degree of freedom 3, the gotten value gives 0.878 i.e, 0.878 or -0.878. Since the value of r of -0.9649 falls within range of -0.878, we reject the null hypothesis. Hence, there is sufficient evidence to support that the linear correlation between x and y exists.
Question c: Yes, the linear correlation exists between x and y. We then construct the least squared line.
[tex]\begin{gathered} y^{\prime}=a+bx \\ \text{where b = slope =}\frac{SS_{xy}}{SS_x} \\ SS_{xy}=-18 \\ SS_x=10 \\ b=\frac{-18}{10}=-1.8 \\ To\text{ get a} \\ a=y-\text{intercept}=\bar{y}-\bar{bx} \\ \bar{y}=\operatorname{mean}\text{ of y values} \\ \bar{y}=9.8 \\ \bar{x}=\operatorname{mean}\text{ of x values} \\ \bar{x}=\frac{15}{3}=5 \\ a=\bar{y}-b\bar{x} \\ a=9.8-(-1.8)(5) \\ a=9.8+9 \\ a=18.8 \end{gathered}[/tex]
Therefore, the least squared line will be:
[tex]\begin{gathered} y^{\prime}=a+b\bar{x} \\ y^{\prime}=18.8+(-1.8)x \\ y^{\prime}=18.8-1.8x \end{gathered}[/tex]
Question d:
The Scattered plot with the line will be:
